[프로그래머스] MYSQL - JOIN

[프로그래머스] MYSQL - JOIN

1. 없어진 기록 찾기

https://school.programmers.co.kr/learn/courses/30/lessons/59042

프로그래머스

SELECT A.ANIMAL_ID,A.NAME
FROM ANIMAL_OUTS A 
LEFT JOIN ANIMAL_INS B
ON A.ANIMAL_ID = B.ANIMAL_ID
WHERE B.ANIMAL_ID IS NULL
ORDER BY ANIMAL_ID

 

2. 있었는데요 없었습니다

https://school.programmers.co.kr/learn/courses/30/lessons/59043

프로그래머스

SELECT A.ANIMAL_ID,A.NAME
FROM ANIMAL_INS AS A
JOIN ANIMAL_OUTS AS B
WHERE A.ANIMAL_ID=B.ANIMAL_ID AND A.DATETIME >= B.DATETIME
ORDER BY A.DATETIME

 

3. 오랜 기간 보호한 동물(1)

https://school.programmers.co.kr/learn/courses/30/lessons/59044

프로그래머스

SELECT NAME, DATETIME
FROM ANIMAL_INS
WHERE ANIMAL_ID NOT IN (
    SELECT ANIMAL_ID
    FROM ANIMAL_OUTS
)
ORDER BY DATETIME
LIMIT 3

 

4. 보호소에서 중성화한 동물

 

https://school.programmers.co.kr/learn/courses/30/lessons/59045

프로그래머스

SELECT A.ANIMAL_ID, A.ANIMAL_TYPE, A.NAME
FROM ANIMAL_INS as A JOIN ANIMAL_OUTS as B
WHERE A.ANIMAL_ID = B.ANIMAL_ID AND A.SEX_UPON_INTAKE !=B.SEX_UPON_OUTCOME
ORDER BY A.ANIMAL_ID